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It is important for you to understand when to use the central limit theorem . If you are being asked to find the probability of the mean, use the clt for the mean. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums.

Note

If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable.

Examples of the central limit theorem

Law of large numbers

The law of large numbers says that if you take samples of larger and larger size from any population, then the mean x ¯ of the sample tends to get closer and closer to μ . From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for X ¯ is σ n .) This means that the sample mean x ¯ must be close to the population mean μ . We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers.

Central limit theorem for the mean and sum examples

A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:

  1. The probability that the mean stress score for the 75 students is less than two.
  2. The 90 th percentile for the mean stress score for the 75 students.
  3. The probability that the total of the 75 stress scores is less than 200.
  4. The 90 th percentile for the total stress score for the 75 students.

Let X = one stress score.

Problems a and b ask you to find a probability or a percentile for a mean . Problems c and d ask you to find a probability or a percentile for a total or sum . The sample size, n , is equal to 75.

Since the individual stress scores follow a uniform distribution, X ~ U (1, 5) where a = 1 and b = 5 (See Continuous Random Variables for an explanation on the uniform distribution).

μ X = a + b 2 = 1 + 5 2 = 3

σ X = ( b a ) 2 12 = ( 5 1) 2 12 = 1.15

For problems 1. and 2., let X ¯ = the mean stress score for the 75 students. Then,

X ¯ N ( 3,  1 .15 75 ) where n = 75.

a. Find P ( x ¯ <2). Draw the graph.

a. P ( x ¯ <2) = 0

The probability that the mean stress score is less than two is about zero.

This is a normal distribution curve over a horizontal axis. The peak of the curve coincides with the point 3 on the horizontal axis. A point, 2, is marked at the left edge of the curve.

normalcdf ( 1,2,3, 1 .15 75 ) = 0

Reminder

The smallest stress score is one.

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b. Find the 90 th percentile for the mean of 75 stress scores. Draw a graph.

b. Let k = the 90 th precentile.

Find k , where P ( x ¯ < k ) = 0.90.

k = 3.2

This is a normal distribution curve. The peak of the curve coincides with the point 3 on the horizontal axis. A point, k, is labeled to the right of 3. A vertical line extends from k to the curve. The area under the curve to the left of k is shaded. The shaded area shows that P(x-bar < k) = 0.90.

The 90 th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2.

invNorm ( 0 .90,3, 1.15 75 ) = 3.2

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For problems c and d, let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N [(75)(3), ( 75 ) (1.15)]

c. Find P ( Σx <200). Draw the graph.

c. The mean of the sum of 75 stress scores is (75)(3) = 225

The standard deviation of the sum of 75 stress scores is ( 75 ) (1.15) = 9.96

P ( Σx <200) = 0

This is a normal distribution curve over a horizontal axis. The peak of the curve coincides with the point 225 on the horizontal axis. A point, 200, is marked at the left edge of the curve.

The probability that the total of 75 scores is less than 200 is about zero.

normalcdf (75,200,(75)(3), ( 75 ) (1.15)).

Reminder

The smallest total of 75 stress scores is 75, because the smallest single score is one.

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Questions & Answers

how to use grouped and ungrouped data
Hassan Reply
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Mano Reply
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saifuddin
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saifuddin
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EGBE
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EGBE
fit a binomial distribution for the following data and test the goodness of fit x: 0 1 2 3 4 5 6 f: 5 18 28 12 7 6 4
Mikki Reply
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It is a square chi
Nelson
But can't be a binomial because, the x numbers are 0 to 6, instead those would be "0" or "1" in a straight way
Nelson
You can do a chi-square test, but the assumption has to be a normal distribution, and the last f's number need to be "64"
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Nelson
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rajendra
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rajendra
Sorry I see my mistake, we have to calculate the expected values
Nelson
So we need this equation: P= (X=x)=(n to x) p^x(1-p)^n-x
Nelson
why it is not possible brother
ibrar
were n= 2 ( binomial) x= number of makes (0 to 6) and p= probability, could be 0.8.
Nelson
so after we calculate the expected values for each observed value (f) we do the chi-square. x^2=summatory(observed-expected)^2 / expected and compare with x^2 in table with 0.8
Nelson
tomorrow I'll post the answer, I'm so tired today, sorry for my mistake in the first messages.
Nelson
It is possible, sorry for my mistake
Nelson
two trader shared investment and buoght Cattle.Mr.Omer bought 255 cows & rented the farm for a period of 32 days. Mr. Ahmed grazed his Cattle for 25 days. Mr. Ahmed's cattle was 180 cows.Together they profited $ 7800. the rent of the farm is $ 3000 so divide the profit per gows/day for grazing day
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People living longer
Qader Reply
Why do you think that is?
Jazzy
because there is an increase in number of people with age more than 30.
Qader
Ok. And what do you think is the driving factor behind that hypothesis?
Jazzy
fewer birth and increase in # of years living or fewer dying
Qader
What about the improvement of technology and medicine?
Jazzy
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Ayunku
If those conscience of their health, one will live longer periods of life.
Montrae
hi,why the mean =sum(xi)/n but the variance =sum(xi-xbar)/ n-1 what is the difference between (n or n-1)
Soran Reply
This is hard to type, so I'll use "m" for "x bar", and a few other notations that I hope will be clear: Definition: sqrt(SUM[(x - m)^2] / (n-1)) where m = SUM[x] / n Desired formula: sqrt((SUM[x^2] - SUM[x]^2)/n / (n-1)) Now let's do what you started to do, and see if we can manipulate the definitio
Michael
what is the difference between (n ) and (n-1) in the mean and variance
Soran
Definition: sqrt(SUM[(x - m)^2] / (n-1)) where m = SUM[x] / n what is the difference between (n and n-1)
Soran
Hi, the diference is tha when we estimate parameters in a sample (not in the total population) we need to consider the degrees of liberty for the estimation.
Nelson
Hie guys, am analysing rainfall data for different stations and i got kurtosis values of 0.7 for one station and 0.4 for another, what can i say about this?
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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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